Suppose $p_0,\ p_1,\ \dots,\ p_q$ are positive such that $p_0+p_1+\dots+p_q=1$. I am wondering how to find the eigenvalues of the following arrowhead matrix $$A=\begin{bmatrix} 1 & p_1 & \dots & p_q\\ p_1 & p_1 & & \\ \vdots & & \ddots & \\ p_q & & & p_q \end{bmatrix} \in \mathbb{R}^{(q+1)\times (q+1)}.$$ where the empty parts in $A$ are all zeros and $p_0$ does not show up in $A$. Since $p_0>0$, we have $0<p_1+\dots+p_q<1$ and $A$ is a positive definite matrix.
I used the method $|\lambda I - A|$ to find the eigenvalues: $$\lambda I - A=\begin{bmatrix} \lambda-1 & -p_1 & \dots & -p_q\\ -p_1 & \lambda-p_1 & & \\ \vdots & & \ddots & \\ -p_q & & & \lambda-p_q \end{bmatrix}.$$ We can multiply the second column of $\lambda I - A$ by $p_1/(\lambda-p_1)$ which could be added to the first column. Then we can eliminate the second entry in the first column. However, this method moves $\lambda-p_1$ in the denominator. It is still hard to find those eigenvalues of $A$.
Let $p=(p_1,p_2,\ldots,p_q)^T$. Using Schur complement, we get $$ \begin{align} \det(\lambda I-A) &=\det\pmatrix{\lambda-1&-p^T\\ -p&\lambda I-\operatorname{diag}(p)}\\ &=\det(\lambda I-\operatorname{diag}(P))\left[(1-\lambda)-p^T(\lambda I-\operatorname{diag})^{-1}p\right]\\ &=\prod_{i=1}^q(\lambda-p_i)\left[(\lambda-1)-\sum_{i=1}^q\frac{p_i^2}{\lambda-p_i}\right]\\ &=(\lambda-1)\prod_{i=1}^q(\lambda-p_i)-\sum_{i=1}^qp_i^2\prod_{\substack{1\le j\le q\\ j\ne i}}(\lambda-p_j)\tag{1}\\ \end{align} $$ Therefore the eigenvalues of $A$ are the zeros of $(1)$. I am not sure if they can be expressed in closed forms.